Answer
$1.8m$
Work Step by Step
We can find the required distance as follows:
$a=\frac{f_K}{m}$
$\implies a=\frac{-\mu_k mg}{m}$
$\implies a=-\mu_k g$
We know that
$\Delta x=\frac{v^2-v_{\circ}^2}{2a}$
$\implies \Delta x=\frac{0-v_{\circ}^2}{-2\mu_K g}$
We plug in the known values to obtain:
$\Delta x=\frac{(4.0)^2}{2(0.46)(9.81)}$
$\Delta x=1.8m$